Joshua Psalms T. answered 06/01/16
Tutor
5
(5)
Civil EIT, Former College Professor of Mathematics (in Asia)
ASSUMING that this is just a simple interest problem and NOT a compounded interest problem:
Let: P1 = Money invested for 6 months; and P2 = Money invested for 1 year.
From this, the first equation is simply the sum of the two which is $410,000:
P1 + P2 = 410000, eq. 1
Next: Let I1 = Interest from P1; and I2 = Interest from P2
The same as the previous one, the sum of the interests is $8761, therefore:
I1 + I2 = 8761, eq. 2
Now, we have two equations but with different sets of variables. However we do know that:
I = Prt, Meaning:
I1 = P1r1t1 = P1(.0262)(0.5) //the 0.5 is the 6 months in terms of years//
I1 = .0131P1
Doing the same with I2, with r2 = 0.0291 and t1 = 1,
I2 = .0291P2
Substituting I1 and I2 to eq. 2, you'll get:
.0131P1 + .0291P2 = 8761 //you can multiply both sides by 100 to make it a little prettier//
1.31P1 + 2.91P2 = 876100, eq. 3
P1 + P2 = 410000, eq. 1 //using eq. 1 and eq. 3, through either substitution or elimination//
P1 = $198125
P2 = $211875